Problem in Proving the potential due to concentric shells between the shells

Imagine a spherical shell having charge Q. The Electric Field exists outside. I bring a test charge from infinity to the surface of the shell (I bring it radially to simplify the calculation and moreover the field is conservative) and solve the Integral to get KQ/a (a is the radius). Now I move the charge inside but there is no field there. Mathematically I divide the Integral in 2 parts , 1st from infinity to a and 2nd from a to b. The Integral from a to b gives 0 and hence the result.

Now imagine two concentric spherical shells of radius a and b. The outer shell carries charge -2Q (radius a) and the inner carries Q. By the above method of bringing the test charge from infinity to the point, I get the potential at the surface of outer sphere as -kQ/a (since field outside is -kQ/(r^2). For the potential at the surface of the inner sphere (by the method above ) , I divide the Integral from infinity to a and from a to b. Now the 1st integral is mathematically just the just the above case (case of potential at the surface of outer sphere) and gives -kQ/a and since the field inside (between the 2 spheres) is only due to Q , I get kQ/b. Thus total potential at b is -kQ/a+ kQ/b. But this is wrong, we know it has to be -2kQ/a + kQ/b.

So where I am going wrong ?

• Title : do you mean the potential difference between the concentric shells? – sammy gerbil Nov 28 '16 at 1:53
• How do you get that the potential difference between a and b is kQ/b? That is the potential difference when you bring the test charge from infinity to b, in the absence of a. Not the potential difference going from a to b. – sammy gerbil Nov 28 '16 at 2:16
• @sammygerbil Yes , I get what you say . It's a problem with sign conventions. Since V= -E.dr , and for the sphere B (inner one) when I bring the test charge from a to b (radially ) my dr becomes -dr. So V becomes -E.(-dr)=E.dr and E is kQ/(r^2) . On integrating , I get kQ/a - kQ/b . I should get kQ/b- kQ/a. That is the the problem with signs. I don't realize where am I missing signs . dr is opposite to E between the spheres so V=E.dr. From somewhere I need a negative sign. Even if for a simply positively charged sphere if I bring a test charge towards it instead of away, I get opposite result – Shashaank Nov 28 '16 at 6:00
• @sammygerbil That was the difficulty with signs. For a positively charged sphere , if I bring a test charges from infinity towards it (radially) rather than away , I get V=-E.dr = -E(-dr)=Edr which gives the opposite result with signs. I guess the entire problem is with signs and interpreting the result ! – Shashaank Nov 28 '16 at 6:05
• – sammy gerbil Nov 28 '16 at 16:56

A solution to this problem is provided on the okphysics website. It does not use integration but assumes that the potential at the surface of an isolated sphere of charge $Q$ and radius $R$ is $kQ/R$. It also uses the superposition theorem and the fact that the charge on each shell can be replaced with a point charge at the centre.

The potential at the surface of the outer shell is that at radius $b$ from charge $Q_a+Q_b$ : ie
$V_b = k\frac{Q_a+Q_b}{b}$.

In going from the outer to the inner shell we are inside charge $Q_b$ so it does not contribute to the potential difference. We move from radius $b$ to radius $a$ from charge $Q_a$. So the potential difference between the 2 shells is
$V_{ab}=kQ_a(\frac{1}{a}-\frac{1}{b})$.

The potential at the surface of the inner shell is
$V_a=V_b+V_{ab}=k(\frac{Q_a}{b}+\frac{Q_b}{b}+\frac{Q_a}{a}-\frac{Q_a}{b})=k(\frac{Q_a}{a}+\frac{Q_b}{b})$

Another way of deriving the last result is by superposition.

First consider the outer shell in isolation. Since there is no electric field inside this shell, the potential is the same at all points inside as it is on the surface : ie $k\frac{Q_b}{b}$. In particular this applies at all points at radius $a<b$ from the centre.

Second consider the inner shell in isolation. The potential at the surface is $k\frac{Q_a}{a}$.

Now superpose the 2 potentials to get the potential at the surface of the inner shell when both charged shells are present :
$V_a=k(\frac{Q_a}{a}+\frac{Q_b}{b})$.

• Thanks , I understood that solution. I wanted the integration one. I will try on my own and if i succeed , I will write my own solution here. – Shashaank Nov 29 '16 at 19:06